Finding a tuple in a tuple

[bg_ie]

Hi,

Lists say I have the following tuple -

t1 = ("ONE","THREE","SIX")

and then the following tuples -

t2 = ("ONE","TWO","THREE")

t3 = ("TWO","FOUR","FIVE","SIX")

t4 = ("TWO",)

t5 = ("TWO","FIVE")

What I want to do is return true if any member of tuple t1 is found in
the remaining tuples.

Therefore -

2) ("ONE","TWO","THREE") : TRUE

3) ("TWO","FOUR","FIVE","SIX") : TRUE

4) ("TWO",) FALSE

5) ("TWO","FIVE")

How do I do this?

Cheers,

Barry.

[Paul Rubin]

bg***@yahoo.com writes:

Lists say I have the following tuple -
t1 = ("ONE","THREE","SIX")
t2 = ("ONE","TWO","THREE")
t3 = ("TWO","FOUR","FIVE","SIX")
t4 = ("TWO",)
t5 = ("TWO","FIVE")

What I want to do is return true if any member of tuple t1 is found in
the remaining tuples.

Convert them into sets and use the set intersection (&) operator,
then convert to bool to represent whether the intersection is empty.

print bool(set(t1) & set(t2))
print bool(set(t1) & set(t3))
print bool(set(t1) & set(t4))
print bool(set(t1) & set(t5))

[Paul McGuire]

On Feb 22, 3:05 am, Paul Rubin <http://[e-mail]:

[e-mail] writes:

Lists say I have the following tuple -
t1 = ("ONE","THREE","SIX")
t2 = ("ONE","TWO","THREE")
t3 = ("TWO","FOUR","FIVE","SIX")
t4 = ("TWO",)
t5 = ("TWO","FIVE")

What I want to do is return true if any member of tuple t1 is found in
the remaining tuples.

Convert them into sets and use the set intersection (&) operator,
then convert to bool to represent whether the intersection is empty.

print bool(set(t1) & set(t2))
print bool(set(t1) & set(t3))
print bool(set(t1) & set(t4))
print bool(set(t1) & set(t5))

A step further: use union to make a superset of t2-tN, then use & on
this superset.

setlist = [t2,t3,t4,t5]
superset = reduce(set.union, map(set,setlist) )
print bool(t1 & superset)
-- Paul

[Paul Rubin]

"Paul McGuire" <pt***@austin.rr.comwrites:

A step further: use union to make a superset of t2-tN, then use & on
this superset.

setlist = [t2,t3,t4,t5]
superset = reduce(set.union, map(set,setlist) )
print bool(t1 & superset)

Well you do have to convert them to sets. Also I thought each
intersection was wanted separately. Otherwise if we're getting this
fancy, I guess I'd use (untested, uses new 2.5 "any" function):

s1 = set(t1)
print any((s1 & set(tn)) for tn in (t2,t3,t4,t5))

which avoids creating a potentially big intermediate set, and which
short-circuits (exits early) as soon as a match is found.

[Philipp Pagel]

bg***@yahoo.com wrote:

t1 = ("ONE","THREE","SIX")
t2 = ("ONE","TWO","THREE")
t3 = ("TWO","FOUR","FIVE","SIX")
t4 = ("TWO",)
t5 = ("TWO","FIVE")

What I want to do is return true if any member of tuple t1 is found in
the remaining tuples.

Another way to go instead of using sets, although probably less elegant:

>>True in [x in t1 for x in t2]

True

>>True in [x in t1 for x in t3]

True

>>True in [x in t1 for x in t4]

False

>>True in [x in t1 for x in t5]

False

cu
Philipp

--
Dr. Philipp Pagel Tel. [phone]
Dept. of Genome Oriented Bioinformatics Fax. [phone]
Technical University of Munich
http://mips.gsf.de/staff/pagel

[Jussi Salmela]

bg***@yahoo.com kirjoitti:

Hi,

Lists say I have the following tuple -

t1 = ("ONE","THREE","SIX")

and then the following tuples -

t2 = ("ONE","TWO","THREE")

t3 = ("TWO","FOUR","FIVE","SIX")

t4 = ("TWO",)

t5 = ("TWO","FIVE")

What I want to do is return true if any member of tuple t1 is found in
the remaining tuples.

Therefore -

2) ("ONE","TWO","THREE") : TRUE

3) ("TWO","FOUR","FIVE","SIX") : TRUE

4) ("TWO",) FALSE

5) ("TWO","FIVE")

How do I do this?

Cheers,

Barry.

Another variation of the theme:

#====================
for t in (t2, t3, t4, t5):
for x in t1:
if x in t:
print True
break
else: print False
#====================
HTH,
Jussi

[Duncan Booth]

Philipp Pagel <pD*******@gsf.dewrote:

Another way to go instead of using sets, although probably less elegant:

>>>True in [x in t1 for x in t2]

True

>>>True in [x in t1 for x in t3]

True

>>>True in [x in t1 for x in t4]

False

>>>True in [x in t1 for x in t5]

False

Slightly more elegant for Python 2.5 users:

>>any(x in t1 for x in t2)

True

>>any(x in t1 for x in t3)

True

>>any(x in t1 for x in t4)

False

>>any(x in t1 for x in t5)

False